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If $\dfrac{dx}{dy} = \sin(x),$ then is $\dfrac{dy}{dx} = \dfrac{1}{\sin(x)}$?

I'm trying to understand how to manipulate $dx$ and $dy$ quantities effectively.

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    $\begingroup$ If Panny walks ten times the speed of Nanny, then does Nanny walk one-tenth of the speed of Panny? $\endgroup$ Dec 14, 2014 at 17:29
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    $\begingroup$ @Venus Could you please provide that link for everyone's benefit? An obvious place to expect an admonishment about "too much vertical space" would be in meta.math.stackexchange.com/questions/9959/… where I see a request to minimize vertical space in titles, but nothing like that for the question itself. $\endgroup$
    – David K
    Dec 15, 2014 at 14:19
  • $\begingroup$ Stop defacing this question. I'm removing unrelated comments. $\endgroup$
    – davidlowryduda
    Dec 15, 2014 at 15:07

1 Answer 1

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By the chain rule (assuming all quantities exist and make sense):

$$\dfrac {\mathrm dx}{\mathrm dy}\dfrac {\mathrm dy}{\mathrm dx} = \dfrac {\mathrm dx}{\mathrm dx} = 1$$

do you see how that answers your question?

edit: the OP asked when this works. This works if $y$ is invertible and the derivative isn't $0$. If $y$ is not invertible, then $x$ might still implicitly define a differentiable function of $y$ for some neighborhood of a point you're interested in.

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  • $\begingroup$ Not really..... :) Where did sin(x) go? $\endgroup$
    – bodacydo
    Dec 15, 2014 at 14:23
  • $\begingroup$ Multiply your two equations together. $\endgroup$ Dec 15, 2014 at 15:10
  • $\begingroup$ It works. :) But can I always do that? For any given $y = f(x)$? $\endgroup$
    – bodacydo
    Dec 15, 2014 at 16:54
  • $\begingroup$ edited my answer to explain when it works $\endgroup$
    – GFauxPas
    Dec 16, 2014 at 13:18
  • $\begingroup$ @bodacydo looks like GFauPax answered your questions. If so, you should accept it so we know its been addressed. $\endgroup$
    – user76844
    Jan 16, 2015 at 6:18

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